What is the tangent of f(x)=x^3+2x ?

The tangent of f(x)=x^3+2x is y=(3a_{0}^2+2)x-2a_{0}^3

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tangent of f(x)=x^3+2x

tangent of

f (x) = x^{3} + 2 x

y = (3 a_{0}^{2} + 2) x - 2 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} + 2 a_{0})

Find the slope of

f (x) = x^{3} + 2 x : \frac{df ( x )}{d x} = 3 x^{2} + 2

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 3 a_{0}^{2} + 2

Find the line with slope m=

3 a_{0}^{2} + 2

and passing through

(a_{0}, a_{0}^{3} + 2 a_{0}) : y = (3 a_{0}^{2} + 2) x - 2 a_{0}^{3}

y = (3 a_{0}^{2} + 2) x - 2 a_{0}^{3}

\frac{\partial}{\partial x} (x e^{x y} + \frac{x}{y ^{2}})

f (x) = x, at x = 64

4 x y y^{^{'}} = 4 y^{2} + 3 x x^{2} + y^{2}

\int (\frac{5}{x} + \frac{12}{x ^{5}}) d x

f (x) = x (x - 1)

What is the tangent of f(x)=x^3+2x ?
The tangent of f(x)=x^3+2x is y=(3a_{0}^2+2)x-2a_{0}^3